2021 AMC 10A Problems/Problem 20

Problem

In how many ways can the sequence be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

Solution (bashing)

We write out the cases. These cases are the ones that work: We count these out and get permutations that work. ~contactbibliophile

Solution 2 (Casework)

Reading the terms from left to right, we have two cases:

(1)

(2)

( stands for increase and stands for decrease.)

For Case (1), note that for the 2nd and 4th terms, one of which must be a 5, and the other one must be a 3 or 4. We have four scenarios:

_3_5_

_5_3_

_4_5_

_5_4_

For the first scenario, the first two blanks must be 1 and 2 in some order, and the last blank must be a 4, for a total of 2 possibilities. Similarly, the second scenario also has 2 possibilities.

For the third scenario, there are no restrictions for the numbers 1, 2, and 3. So, we have possibilities. Similarly, the fourth scenario also has 6 possibilities.

Together, Case (1) has possibilities. By symmetry, Case (2) also has possibilities. Together, the answer is

This problem is a little similar to the 2004 AIME I Problem 6: https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6

~MRENTHUSIASM

Solution 3 (similar to solution 2)

Like Solution 2, we have two cases. Due to symmetry, we just need to count one of the cases. For the purpose of this solution, we will be doing . Instead of starting with 5, we start with 1.

There are two ways to place it:

_1_ _ _

_ _ _1_

Now we place 2, it can either be next to 1 and on the outside, or is place in where 1 would go in the other case. So now we have another two "sub case":

_1_2_(case 1)

21_ _ _(case 2)

There are 3! ways to arrange the rest for case 1, since there is no restriction.

For case 2, we need to consider how many ways to arrange 3,4,5 in a a>b<c fashion. It should seem pretty obvious that b has to be 3, so there will be 2! way to put 4 and 5.

Now we find our result, times 2 for symmetry, times 2 for placement of 1 and times (3!+2!) for the two different cases for placement of 2. This give us .

Video Solution by OmegaLearn (Using PIE - Principal of Inclusion Exclusion)

https://youtu.be/Fqak5BArpdc

~ pi_is_3.14

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.